Rearrange elements of matrix based on an index matrix

2 次查看（过去 30 天）
Hossein Kazemi
Hossein Kazemi 2024-8-27
评论： Hossein Kazemi 2024-8-27
Ran in:
I have a 5x3 matrix and I want to rearrange each row according to the correponding row of a 5x3 index matrix
x=randn(5,3)
x = 5x3
-0.2616 0.3522 -0.4451 -1.1699 0.8921 1.7676 1.4354 1.1977 -0.3197 -1.2002 0.9392 0.7452 -0.8374 -1.3827 0.2679
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
z=randn(5,3)
z = 5x3
-0.1687 0.4662 1.3644 0.1290 -3.2728 -0.8944 -0.7820 0.2506 -1.0295 0.0976 -0.0797 -0.2671 0.2607 -0.0966 -0.1679
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[~,I]=sort(x,2)
I = 5x3
3 1 2 1 2 3 3 2 1 1 3 2 2 1 3
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Now I want to sort rows of z using the index matrix I. But using the following does not work. For example, I want the first row of zz to be sorted according to the first row of x, which should result in zz(1,:)= [1.3644, -0.1687, 0.4662].
zz=z(I)
zz = 5x3
-0.7820 -0.1687 0.1290 -0.1687 0.1290 -0.7820 -0.7820 0.1290 -0.1687 -0.1687 -0.7820 0.1290 0.1290 -0.1687 -0.7820
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

请先登录，再回答此问题。

采纳的回答

Stephen23
Stephen23 2024-8-27
编辑：Stephen23 2024-8-27
Ran in:
Yes, it is awkward.
x=randn(5,3)
x = 5x3
0.9687 -0.7929 2.0516 -0.1236 0.2917 -0.9487 -1.4717 0.0275 1.1722 0.2800 1.5357 0.2596 -0.9693 -0.1854 0.3145
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
z=randn(5,3)
z = 5x3
-1.6696 0.6182 -1.4586 0.1528 1.6729 1.4683 0.3576 -1.0089 0.8616 2.1993 0.6213 -1.2327 -0.3246 -0.8133 1.0463
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[~,I] = sort(x,2)
I = 5x3
2 1 3 3 1 2 1 2 3 3 1 2 1 2 3
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Perhaps
S = size(I);
[R,~] = ndgrid(1:S(1),1:S(2));
J = sub2ind(S,R,I);
zz = z(J)
zz = 5x3
0.6182 -1.6696 -1.4586 1.4683 0.1528 1.6729 0.3576 -1.0089 0.8616 -1.2327 2.1993 0.6213 -0.3246 -0.8133 1.0463
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Or
zz = z;
for k = 1:size(I,1)
zz(k,:) = zz(k,I(k,:));
end
zz
zz = 5x3
0.6182 -1.6696 -1.4586 1.4683 0.1528 1.6729 0.3576 -1.0089 0.8616 -1.2327 2.1993 0.6213 -0.3246 -0.8133 1.0463
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Or
zz = cell2mat(cellfun(@(v,x)v(x),num2cell(z,2),num2cell(I,2),'uni',0))
zz = 5x3
0.6182 -1.6696 -1.4586 1.4683 0.1528 1.6729 0.3576 -1.0089 0.8616 -1.2327 2.1993 0.6213 -0.3246 -0.8133 1.0463
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Operating on Diagonal Matrices 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by